The graph of the function f(x)=ax^2+bx+c (where a, b, and c are real and nonzero has two x- intercepts. Explain how to find the other x-intercept if one x-intercept is at ((-b)/(2a)+3,0) square

To find the other x-intercept of the quadratic function \( f(x) = ax^2 + bx + c \), given that one x-intercept is at \( \left( \frac{-b}{2a} + 3, 0 \right) \), we can use the fact that the x-intercepts (roots) of the quadratic equation are symmetric about the vertex of the parabola.<br /><br />The vertex of the parabola given by \( f(x) = ax^2 + bx + c \) is at \( \left( \frac{-b}{2a}, f\left( \frac{-b}{2a} \right) \right) \). Since the given x-intercept is \( \left( \frac{-b}{2a} + 3, 0 \right) \), we know that this point is 3 units to the right of the vertex.<br /><br />The x-coordinate of the vertex is \( \frac{-b}{2a} \). Therefore, the x-coordinate of the given x-intercept is \( \frac{-b}{2a} + 3 \).<br /><br />Since the parabola is symmetric about the vertex, the other x-intercept will be the same distance to the left of the vertex as the given x-intercept is to the right of the vertex. Thus, the other x-intercept will be at:<br /><br />\[ \left( \frac{-b}{2a} - 3, 0 \right) \]<br /><br />So, the other x-intercept of the quadratic function \( f(x) = ax^2 + bx + c \) is at \( \left( \frac{-b}{2a} - 3, 0 \right) \).